Question

Correct formula for height of a satellite from earths surface is :

• A. \( \left( \frac{T^2 R^2 g}{4 \pi} \right)^{1/2} - R \)
• B. \( \left( \frac{T^2 R^2 g}{4 \pi^2} \right)^{1/3} - R \)
• C. \( \left( \frac{T^2 R^2}{4 \pi^2 g} \right)^{1/3} - R \)
• D. \( \left( \frac{T^2 R^2}{4 \pi^2 g} \right)^{-1/3} + R \)

Answer 1

The Correct Option is B

1. Gravitational Force Equals Centripetal Force:
The gravitational force on a satellite of mass m orbiting Earth at height h is equal to the centripetal force required for its orbit:
GMm / (R + h)² = mv² / (R + h).
This simplifies to:
GM / (R + h) = v². (1)

2. Orbital Velocity and Period Relationship:
The orbital velocity v is also defined by the orbital period T as:
v = 2π(R + h) / T. (2)

3. Equating Gravitational Force with Centripetal Acceleration:
Using the relation GM = gR² (where g is Earth's surface gravity), substitute into equation (1):
gR² / (R + h) = v².

4. Combining Equations (1) and (2):
Substitute v from equation (2) into the expression from step 3:
gR² / (R + h) = (2π(R + h) / T)².
Rearranging yields:
T²R²g / (2π)² = (R + h)³.

5. Solving for Height h:
Take the cube root of both sides:
R + h = (T²R²g / 4π²)^(1/3).
Therefore, the height h is:
h = (T²R²g / 4π²)^(1/3) - R.

Answer: (T²R²g / 4π²)^(1/3) - R (Option 2)